Why STP Must Be Defined in HSC Gas Calculations

One mole of an ideal gas occupies 22.71 L at 0°C and 100 kPa. At 25°C and 100 kPa, the same mole occupies 24.79 L.

The amount of gas has not changed. So why is the volume different?

This is exactly why standard temperature and pressure (STP) must be defined in a gas question. Gas volume depends on both temperature and pressure. Unless these conditions are specified, two gas volumes cannot be compared fairly, and you cannot reliably use volume to determine the amount of gas.

There is another complication: different chemistry references have historically used different definitions of STP. A value of 22.4 L per mole is not necessarily wrong, but it does not use the same reference pressure as 22.71 L per mole.

Let’s work through where these differences come from, how scientists account for them in experiments, and what to do when an HSC question gives you unfamiliar gas conditions.

01Why gas volume needs a reference condition

Imagine filling two identical gas syringes with the same amount of nitrogen.

You place one in an ice bath and leave the other in a warm room. Both pistons can move freely, and the external pressure is the same.

Which syringe will have the larger volume?

The warmer one.

Its gas particles have greater average kinetic energy, so they move faster. The gas expands against the movable piston until its pressure balances the external pressure.

The number of particles has not changed. Their average motion has.

Pressure matters as well. If you increase the external pressure, the piston is pushed inward, reducing the volume of the gas.

This gives us two important relationships for a fixed amount of gas:

• Increasing temperature increases volume when pressure is constant.
• Increasing pressure decreases volume when temperature is constant.

The precise relationship is expressed by the ideal gas equation:

\[ PV=nRT \]

Where:

Notice that temperature is measured in kelvin, not degrees Celsius.

Convert Celsius to kelvin using:

\[ T(\mathrm K)=T(°\mathrm C)+273.15 \]

For example:

• 0°C = 273.15 K
• 25°C = 298.15 K
• 100°C = 373.15 K

You cannot substitute 25 directly into the ideal gas equation. The temperature must be measured from absolute zero, which is why kelvin is used.

The equation also explains why a reference condition is necessary. For a fixed amount of gas:

\[ V\propto\frac{T}{P} \]

A volume measurement is meaningful only when its temperature and pressure are known.

02What exactly does STP mean?

Standard temperature and pressure provide a common reference point for comparing gas measurements.

However, there is an important detail: STP has not always meant exactly the same thing in every chemistry reference.

The current IUPAC definition of standard conditions for gases is:

Current IUPAC convention

• Temperature: 273.15 K (0°C)
• Pressure: 100 kPa (100,000 Pa or 1 bar)

IUPAC recommends this pressure rather than the older convention of 1 atmosphere.

Source: IUPAC Gold Book – STP and standard conditions for gases.

Why do some textbooks use 22.4 L instead of 22.71 L?

This is a common source of confusion.

Older chemistry references frequently defined STP using a pressure of 1 atmosphere rather than 100 kPa.

These pressures are close, but not identical:

\[ 1\text{ atm}=101.325\text{ kPa} \]

The difference changes the volume occupied by one mole of gas.

Using the ideal gas equation, the molar volume is:

\[ V_m=\frac{RT}{P} \]

Here, \(V_m\) is the volume occupied by one mole of gas.

The following table compares the different reference conditions.

The 22.4 L value is simply the rounded molar volume at 0°C and 1 atmosphere.

The 22.71 L value uses 0°C and 100 kPa.

Neither value describes a different chemical property. They describe the same ideal gas under different reference pressures.

NESA’s Chemistry Stage 6 data sheet gives the molar volume of an ideal gas as 22.71 L at 0°C and 100 kPa, and 24.79 L at 25°C and 100 kPa.

You can find the reference values in the NESA Chemistry formulae sheet and data sheet.

The important exam rule: Never assume that the number 22.4 L mol⁻¹ is appropriate just because a question mentions STP. Use the temperature and pressure specified in the question or the convention supplied by the relevant data sheet.

If the question does not define its convention, state the assumption you are using.

03Using the ideal gas equation with the correct units

Knowing the standard conditions is only half the job. You also need to make sure your units match the gas constant.

The ideal gas constant, \(R\), can be expressed in several equivalent forms.

For HSC Chemistry, the most convenient combination is usually:

• Pressure in kPa
• Volume in L
• Temperature in K
• Amount in mol
• \(R=8.314\) kPa L mol⁻¹ K⁻¹

You may notice that the NESA data sheet gives the gas constant as 8.314 J mol⁻¹ K⁻¹.

These values are numerically identical because:

\[ 1\text{ kPa}\times1\text{ L}=1\text{ J} \]

This is why you can use 8.314 when pressure is in kPa and volume is in L.

For a more detailed explanation of matching pressure, volume, and gas constant units, see Choosing R When Pressure Is in kPa and Volume Is in L.

An important detail: pressure must be absolute

The pressure in the ideal gas equation is absolute pressure, not gauge pressure.

Absolute pressure is measured relative to a complete vacuum.

Gauge pressure is measured relative to the surrounding atmospheric pressure.

For a positive gauge pressure:

\[ P_{\mathrm{absolute}}=P_{\mathrm{gauge}}+P_{\mathrm{atmospheric}} \]

For example, if a pressure gauge reads 200 kPa and atmospheric pressure is 100 kPa, the absolute pressure is:

\[ P_{\mathrm{absolute}}=200+100 \]

\[ P_{\mathrm{absolute}}=300\text{ kPa} \]

Using 200 kPa directly in the ideal gas equation would give an incorrect amount of gas.

This distinction becomes particularly important when interpreting experimental measurements.

04What happens when a gas is collected over water?

Consider a common school laboratory experiment.

Magnesium reacts with hydrochloric acid to produce hydrogen gas. You want to measure the volume of hydrogen produced.

A convenient method is to collect the gas over water using an inverted, graduated measuring tube.

The measurement seems straightforward:

1. Magnesium reacts with hydrochloric acid.
2. Hydrogen travels through the delivery tube.
3. Hydrogen enters the inverted measuring tube and displaces water.
4. You read the volume from the graduated scale.

But there is a problem.

The gas collected is not pure hydrogen.

Some water evaporates into the measuring tube, so the collected gas contains both hydrogen and water vapour.

This means the measured pressure belongs to a mixture of gases.

Why the measured pressure is not the hydrogen pressure

Suppose you adjust the water levels inside and outside the measuring tube until they are equal.

The pressure inside the tube is now equal to atmospheric pressure.

However, that pressure comes from two components:

• Hydrogen gas
• Water vapour

Dalton’s law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of its individual gases.

Therefore:

\[ P_{\mathrm{total}}=P_{\mathrm{H_2}}+P_{\mathrm{H_2O}} \]

Rearranging:

\[ P_{\mathrm{H_2}}=P_{\mathrm{total}}-P_{\mathrm{H_2O}} \]

Since the water levels are equal:

\[ P_{\mathrm{total}}=P_{\mathrm{atmospheric}} \]

Therefore:

\[ P_{\mathrm{H_2}}= P_{\mathrm{atmospheric}}-P_{\mathrm{H_2O}} \]

This is an experimental correction.

You must subtract the water vapour pressure to determine the pressure of the dry hydrogen.

Notice that this correction is not needed because the ideal gas equation is inaccurate. It is needed because the measured gas contains more than one substance.

You must first identify the pressure contributed by the gas you actually want to measure.

If the water levels are not equal, the pressure inside the measuring tube also differs from atmospheric pressure because of the water-column height. An additional hydrostatic pressure correction is then needed.

What does temperature have to do with water vapour?

Water vapour pressure increases with temperature.

For example, the approximate saturated water vapour pressures are:

These are illustrative reference values for the calculations in this article.

At higher temperatures, more water molecules enter the gas phase, increasing the contribution of water vapour to the total pressure.

This is another reason why recording the temperature of an experimental gas sample matters.

05Ideal gas assumptions versus experimental corrections

There are two separate ideas that students sometimes mix up.

The first is the ideal gas model. The second is correcting measurements so that they accurately represent the gas being studied.

1. The ideal gas model

The ideal gas equation assumes that:

• Gas particles occupy negligible volume compared with the container.
• There are no significant attractive or repulsive forces between particles.
• Collisions between particles and container walls are perfectly elastic.
• Gas particles move continuously and randomly.

These assumptions allow us to use:

\[ PV=nRT \]

Real gases do not behave perfectly this way.

At high pressures, particle volume becomes more significant. At low temperatures, intermolecular attractions become more important.

Real gases can therefore deviate from ideal behaviour, particularly near condensation.

The ideal gas equation is an approximation that works well under suitable conditions. STP does not magically make a real gas perfectly ideal.

2. Experimental corrections

Experimental corrections deal with the conditions under which measurements are taken.

These corrections do not all fix the same problem.

For example, subtracting water vapour pressure does not correct for non-ideal gas behaviour. It simply identifies the pressure contributed by the dry gas.

Likewise, converting Celsius to kelvin does not change the physical state of the gas. It changes the temperature scale so that the equation can be used correctly.

For standard HSC ideal gas calculations, assume ideal behaviour unless the question provides information requiring a different model.

06Worked examples

07Questions and solutions

The following questions move from standard gas calculations to more demanding situations involving measurement corrections, competing STP conventions, and changing gas mixtures.

All experimental measurements identified below are illustrative.

Question 1

A sample contains 0.0850 mol of helium gas.

Calculate the volume occupied by the gas at STP, defined as 0°C and 100 kPa.

Assume ideal gas behaviour.

Solution 1

The helium occupies 1.93 L at STP.

The ideal gas equation is:

\[ PV=nRT \]

Rearranging:

\[ V=\frac{nRT}{P} \]

The known values are:

• \(n=0.0850\) mol
• \(T=273.15\) K
• \(P=100\) kPa
• \(R=8.314\) kPa L mol⁻¹ K⁻¹

Substituting:

\[ V=\frac{0.0850\times8.314\times273.15}{100} \]

\[ \boxed{V=1.93\text{ L}} \]

The answer is consistent with the molar volume of 22.71 L mol⁻¹ at these conditions.

The common mistake is to substitute 0 directly for temperature or to use 22.4 L mol⁻¹ without checking the pressure convention.

Both approaches would give an inconsistent result.

Question 2

A sealed rigid cylinder has a volume of 2.00 L and contains an ideal gas at 27.0°C.

A pressure gauge attached to the cylinder reads 250 kPa. This is gauge pressure, not absolute pressure.

The atmospheric pressure is 100 kPa.

Calculate the amount of gas inside the cylinder in moles.

Explain why using 250 kPa directly in the ideal gas equation would produce an incorrect answer.

Solution 2

The cylinder contains 0.281 mol of gas.

The important step is converting gauge pressure to absolute pressure before using the ideal gas equation.

A gauge measures pressure relative to the surrounding atmosphere.

Therefore:

\[ P_{\mathrm{absolute}}= P_{\mathrm{gauge}}+P_{\mathrm{atmospheric}} \]

\[ P_{\mathrm{absolute}}=250+100 \]

\[ P_{\mathrm{absolute}}=350\text{ kPa} \]

Next, convert the temperature:

\[ T=27.0+273.15 \]

\[ T=300.15\text{ K} \]

Use:

\[ PV=nRT \]

Rearrange:

\[ n=\frac{PV}{RT} \]

Substitute:

\[ n= \frac{350\times2.00} {8.314\times300.15} \]

\[ \boxed{n=0.281\text{ mol}} \]

Why is using 250 kPa incorrect?

The gauge reading measures pressure above atmospheric pressure. It does not measure the total pressure exerted by the gas relative to a vacuum.

Using 250 kPa directly would give approximately 0.200 mol, an underestimate of the actual amount.

The ideal gas equation requires absolute pressure, regardless of whether the pressure reading comes from a gauge, a barometer, or another instrument.

Question 3

Two chemistry reference books give different values for the molar volume of an ideal gas at STP.

• Book A gives 22.4 L mol⁻¹ at 0°C.
• Book B gives 22.71 L mol⁻¹ at 0°C and 100 kPa.

A student concludes that one book must be incorrect.

(a) Explain why both values can be correct.

(b) Use the ideal gas equation to estimate the pressure assumed by Book A.

(c) Calculate the percentage difference between the two quoted molar volumes, using 22.71 L mol⁻¹ as the reference value.

(d) Which value should be used to calculate the volume of one mole of an ideal gas at 0°C and 100 kPa?

Solution 3

Both values can be correct because they use different reference pressures. Book A uses approximately 1 atmosphere, whereas Book B uses 100 kPa.

(a) Explain the difference.

The older convention for STP uses:

\[ T=273.15\text{ K} \]

\[ P=101.325\text{ kPa} \]

The current IUPAC convention uses:

\[ T=273.15\text{ K} \]

\[ P=100\text{ kPa} \]

Since volume depends on pressure, the molar volumes are different.

(b) Calculate the pressure assumed by Book A.

For one mole:

\[ V=22.4\text{ L} \]

Rearrange the ideal gas equation:

\[ P=\frac{nRT}{V} \]

Substitute:

\[ P=\frac{1\times8.314\times273.15}{22.4} \]

\[ \boxed{P\approx101.4\text{ kPa}} \]

This is approximately 1 atmosphere (101.325 kPa).

The small difference comes from rounding 22.41 L to 22.4 L.

(c) Calculate the percentage difference.

Using 22.71 L mol⁻¹ as the reference:

\[ \%\text{ difference}= \frac{|22.71-22.4|}{22.71}\times100 \]

\[ \boxed{\%\text{ difference}\approx1.37\%} \]

The difference is relatively small, but it is large enough to produce an inconsistent answer in a precise gas calculation.

(d) Select the correct molar volume.

For 0°C and 100 kPa, use:

\[ \boxed{V_m=22.71\text{ L mol}^{-1}} \]

The conditions in the question determine which value is appropriate.

The tempting approach is to memorise one molar volume and use it everywhere. The correct approach is to identify both temperature and pressure before selecting a reference value.

Question 4

A student reacts 0.300 g of magnesium with excess hydrochloric acid.

The hydrogen gas produced is collected over water in a graduated gas burette.

The following illustrative measurements are recorded:

The water levels inside and outside the burette are equal.

The balanced chemical equation is:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

(a) Calculate the partial pressure of hydrogen.

(b) Calculate the experimental amount of hydrogen collected.

(c) Calculate the theoretical amount of hydrogen that could be produced from the magnesium.

(d) Determine the percentage yield of hydrogen.

Assume ideal gas behaviour and that the measured dry gas is hydrogen.

Solution 4

The experimental amount of hydrogen is \(9.54\times10^{-3}\) mol, giving a percentage yield of 77.3%.

The key is to correct the measured pressure before calculating the amount of hydrogen.

(a) Calculate the partial pressure of hydrogen.

The gas is collected over water, so its total pressure includes water vapour.

Using Dalton’s law:

\[ P_{\mathrm{H_2}}= P_{\mathrm{total}}-P_{\mathrm{H_2O}} \]

Substitute:

\[ P_{\mathrm{H_2}}=100.4-4.24 \]

\[ \boxed{P_{\mathrm{H_2}}=96.16\text{ kPa}} \]

(b) Calculate the experimental amount of hydrogen.

Convert the volume to litres:

\[ V=250.0\text{ mL}=0.2500\text{ L} \]

Convert the temperature to kelvin:

\[ T=30.0+273.15 \]

\[ T=303.15\text{ K} \]

Use the ideal gas equation:

\[ n=\frac{PV}{RT} \]

Substitute:

\[ n= \frac{96.16\times0.2500} {8.314\times303.15} \]

\[ \boxed{n(\mathrm{H_2})=0.00954\text{ mol}} \]

This is the experimental amount of hydrogen collected.

(c) Calculate the theoretical amount of hydrogen.

First, calculate the amount of magnesium:

\[ n=\frac{m}{M} \]

Where \(m\) is mass and \(M\) is molar mass.

\[ n(\mathrm{Mg})=\frac{0.300}{24.31} \]

\[ n(\mathrm{Mg})=0.01234\text{ mol} \]

From the balanced equation:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

The mole ratio of magnesium to hydrogen is 1:1.

Therefore:

\[ n_{\mathrm{theoretical}}(\mathrm{H_2}) =0.01234\text{ mol} \]

(d) Calculate the percentage yield.

\[ \%\text{ yield}= \frac{\text{actual yield}}{\text{theoretical yield}} \times100 \]

Substitute:

\[ \%\text{ yield}= \frac{0.00954}{0.01234}\times100 \]

\[ \boxed{\%\text{ yield}=77.3\%} \]

The experimental yield is lower than the theoretical yield.

Possible explanations include incomplete reaction, gas escaping through leaks, or hydrogen being lost during collection.

The important point is that using atmospheric pressure directly as the hydrogen pressure would overestimate the amount of hydrogen collected.

The water vapour correction must happen before the experimental yield is calculated.

Question 5

A student collects a sample of gas A over water in a graduated gas burette.

The gas is insoluble in water, does not react with water, and no gas escapes during the experiment.

There is excess liquid water in the apparatus, so the gas remains saturated with water vapour throughout the experiment.

The following initial conditions are recorded:

The student then warms the apparatus to 40.0°C.

At the final temperature, the water levels are adjusted to be equal again. Atmospheric pressure remains constant at 100.0 kPa.

The water vapour pressure at 40.0°C is 7.38 kPa.

The student argues:

“The pressure is 100.0 kPa at both temperatures, so the combined gas law tells us that the final volume is simply the initial volume multiplied by the ratio of the temperatures.”

(a) Explain why this prediction is tempting.

(b) Calculate the correct final volume of gas A at 40.0°C.

(c) Calculate the volume predicted by the student’s method.

(d) Explain precisely why the two answers differ, even though the total pressure remains constant.

Assume ideal behaviour for both gases and that gas A does not dissolve, escape, or react.

Solution 5

The correct final volume is 112.6 mL, not the 106.8 mL predicted by the student’s method.

The student’s mistake is assuming that the total amount of gas in the burette remains constant.

It does not.

The amount of gas A remains constant, but the amount of water vapour in the gas mixture increases as the temperature rises.

This distinction changes the calculation.

(a) Explain why the prediction is tempting.

The student recognises that:

• The total pressure is 100.0 kPa at both temperatures.
• The temperature increases.
• The gas volume should increase when temperature increases at constant pressure.

For a fixed amount of ideal gas, the combined gas law gives:

\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]

Since the pressures are equal, this appears to simplify to:

\[ \frac{V_2}{V_1}=\frac{T_2}{T_1} \]

The reasoning is correct for a fixed amount of gas.

The problem is that the total amount of gas in this experiment is not fixed.

(b) Calculate the correct final volume.

We must consider the partial pressure of gas A, rather than the total pressure of the mixture.

First, calculate the initial partial pressure of gas A.

\[ P_{\mathrm{A,1}}= 100.0-2.34 \]

\[ P_{\mathrm{A,1}}=97.66\text{ kPa} \]

Now calculate the final partial pressure of gas A.

\[ P_{\mathrm{A,2}}= 100.0-7.38 \]

\[ P_{\mathrm{A,2}}=92.62\text{ kPa} \]

The amount of gas A is constant because it does not dissolve, escape, or react.

We can therefore apply the combined gas law to gas A alone:

\[ \frac{P_{\mathrm{A,1}}V_1}{T_1} = \frac{P_{\mathrm{A,2}}V_2}{T_2} \]

Convert both temperatures to kelvin:

\[ T_1=293.15\text{ K} \]

\[ T_2=313.15\text{ K} \]

Rearrange:

\[ V_2= \frac{P_{\mathrm{A,1}}V_1T_2} {P_{\mathrm{A,2}}T_1} \]

Substitute:

\[ V_2= \frac{97.66\times100.0\times313.15} {92.62\times293.15} \]

\[ \boxed{V_2=112.6\text{ mL}} \]

(c) Calculate the student’s predicted volume.

The student uses the total pressure at both temperatures.

Since both total pressures are 100.0 kPa, the calculation becomes:

\[ V_2=V_1\frac{T_2}{T_1} \]

Substitute:

\[ V_2=100.0\times\frac{313.15}{293.15} \]

\[ \boxed{V_2=106.8\text{ mL}} \]

This is the incorrect prediction.

(d) Explain why the answers differ.

The crucial difference is the behaviour of water vapour.

As the temperature increases, the saturated water vapour pressure increases from 2.34 kPa to 7.38 kPa.

More water evaporates into the gas phase.

Consequently:

• The amount of gas A remains constant.
• The amount of water vapour increases.
• The total amount of gas in the burette increases.
• The total pressure remains equal to atmospheric pressure because the water levels are equal.

The student’s calculation treats the entire gas mixture as though it contains a fixed number of moles.

That assumption is incorrect.

The correct calculation uses the partial pressure of gas A, because it is the amount of gas A that remains constant.

This is a much more subtle issue than simply remembering to convert Celsius to kelvin or subtract water vapour pressure.

The combined gas law only applies directly when the amount of the gas being considered remains constant.

That condition matters more than whether the total pressure happens to remain unchanged.

08What to remember when answering an HSC gas question

Before substituting values into the ideal gas equation, establish exactly what gas you are calculating for and under which conditions it is being measured.

A reliable sequence is:

1. Identify the temperature and pressure specified in the question.
2. Check whether STP has been explicitly defined.
3. Convert temperature into kelvin.
4. Make sure pressure is absolute, not gauge pressure.
5. Check whether the gas is mixed with water vapour or another gas.
6. Apply any necessary experimental pressure corrections.
7. Choose an appropriate value of \(R\) for your units.
8. Use the ideal gas equation or combined gas law, depending on whether the amount of gas remains constant.

Remember three particularly important distinctions:

• 22.4 L mol⁻¹ and 22.71 L mol⁻¹ use different standard pressure conventions.
• A gas collected over water has a total pressure that includes water vapour.
• The combined gas law requires a fixed amount of the gas being considered.

For NSW students in 2026-27, the Chemistry Stage 6 Syllabus (2017) remains the taught course; the 2025 Chemistry 11-12 syllabus begins implementation for Year 11 in 2028.

Once you can identify the correct temperature, pressure, and amount of gas, the next useful step is understanding how these variables change together.

In particular, learn to distinguish a gas undergoing a change of state from a gas being produced or consumed in a chemical reaction. That decision determines whether you should use the combined gas law or the full ideal gas equation.

For the graphical side of these relationships, continue with How to Read Pressure-Volume and Temperature Gas Graphs.